$A, B, C$ are finitely generated $F$-modules. Where $F$ is PID.

  1. Prove that if $A \hookrightarrow B, B \hookrightarrow A$, then $A\cong B$.
  2. If $A \bigoplus B \cong A \bigoplus C$, is it true $B \cong C$ ?

As I understood in first problem I should prove that this injection is also a surjection. In second problem I don't know where to start.

  • 2
    $\begingroup$ So $A\to B$, $B\to A$ injections give you $0\to A\to B\to B/A\to 0$ split exact sequence. In particular, this says $B\cong A\oplus B/A$. $B$ is embedding in $A$. Says exactly $B/A=0$. So you conclude that $B\cong A$. For 2, consider $(A\oplus B)/A\cong B$ and $(A\oplus C)/A\cong C$. $\endgroup$ – user45765 Sep 10 '17 at 18:30
  • $\begingroup$ @user45765 For the exact sequence to be split, you would need that the composition of the injections $A\to B$ and $B\to A$ is the identity on $A$. This does not seem to follow from the hypothesis: take $A=B=\mathbb{Z}$ and let both inclusions be multiplication by $2$. $\endgroup$ – Pierre-Guy Plamondon Sep 10 '17 at 19:06
  • $\begingroup$ @Pierre-GuyPlamondon Say $i:A\to B, j:B\to A$. I can check $Id_A-j\circ i:A\to A$ as $0$ map. So $Id_A=j\circ i$ on every $a\in A$. Thus it is a split. $\endgroup$ – user45765 Sep 10 '17 at 19:13
  • 1
    $\begingroup$ @user45765 If $i$ and $j$ are multiplication by $2$ in $\mathbb{Z}$, then $j\circ i$ is multiplication by $4$. This is not the identity. $\endgroup$ – Pierre-Guy Plamondon Sep 10 '17 at 19:14
  • $\begingroup$ @Pierre-GuyPlamondon Ah my bad. I thought naively that these are embeddings. In general, they are not. I will think about this. Thanks. $\endgroup$ – user45765 Sep 10 '17 at 19:26

The argument for 1 is quite intricate. Over PID, a submodule of free module is free and ranks is strictly smaller than that of free module. I will take $A\to B$ as embedding and $i:B\to A$ as a different injection. $F_X$ denotes free module surjective onto the generators of module $X$ and $K_X$ denotes the corresponding kernel.

$0\to K_A\to F_A\to A\to 0$

$0\to K_B\to F_B\to B\to 0$

$K_A$ is a submodule of free module and hence free. First use $A\to B$ to induce a morphism from $f_F:F_A\to F_B$ by projectivity. This induces morphism $f_K:K_A\to K_B$. From snake lemma, we have $0\to Ker(f_F)\to Ker(f_K)\to 0\to\dots$ where $\dots$ part I do not care. In particular, this says $Ker(f_F)$ is free and the morphism is restriction of $K_A\to F_A$. You can remove redundant part of $F_A$ and assume $f_F:F_A\to F_B$ and $f_K:K_A\to K_B$ injective. Now $f_F$ says $F_A$'s image is a submodule of $F_B$ which is free. In particular, the rank of $F_A$ must be strictly smaller than that of $F_B$. Now reverse argument by $i:B\to A$. You will see $F_A$'s rank is exactly that of $F_B$. Similarly for $K_A,K_B$. You can induce obvious morphism $K_A\to K_B$ and $F_A\to F_B$ by identity morphism as both $K_X,F_X$ are free for $X=A,B$. Thus $A\cong B$ from snake lemma again.

Unfortunately, my incapacity of figuring out a simpler argument. There should be some better argument to make it simpler proof.

The simpler proof of 1 comes from using chinese remainder theorem.

For 2, use PID module decomposition into torsion free part and torsion part. Then use chinese remainder theorem to see the rest.


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